Since that
\(e_{k}^{\mu }(x,u)=e^{-i\mu (u)}\left( \varphi _{k}^{\left|\mu \right|}\circ A_{\frac{\mu }{\left|\mu \right|}}\right)(x)\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|,\) 
we have
\(S^{\delta }(x,u) &=&\frac{1}{(2\pi )^{n+m}}\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}\left( 1-(2k+n)\left|\mu \right|\right) _{+}^{\delta }e_{k}^{\mu }(x,u)\left|\mu \right|^{n}d\mu \\&=&\frac{1}{(2\pi )^{n+m}}\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}e^{-i\mu (u)}\left( 1-(2k+n)\left|\mu \right|\right)_{+}^{\delta }\varphi _{k}^{\left|\mu \right|}\left( A_{\frac{\mu }{\left|\mu \right|}}x\right) \left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\left|\mu \right|^{n}d\mu .\) 
Let \(z=A_{\frac{\mu }{\left|\mu \right|}}x\)  and
\(F(z,u)=\frac{1}{(2\pi )^{n+m}}\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}e^{-i\mu (u)}\left( 1-(2k+n)\left|\mu \right|\right)_{+}^{\delta }\varphi _{k}^{\left|\mu \right|}\left( z\right)\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\left|\mu \right|^{n}d\mu . \) 
\(F(z,u)\)  is a radial fuction with respect to \(z\) . We set \(r=\left|z\right|\)  and define
\(R_{k}(\mu ,F)=\frac{2^{(1-n)}k!}{(k+n-1)!}\int _{0}^{\infty }F^{\mu }(r)\varphi _{k}^{\left|\mu \right|}(r)r^{2n-1}dr, \) 
where
\(F^{\mu }(r)=\int _{\mathbb {R}^{m}}F(r,u)e^{i\mu (u)}du.\) 
Using the inversion formula of the Fourier transform and the Laguerre
expension of the radial function (see [1]}), we have that
\(F(z,u)=\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}e^{-i\mu (u)}R_{k}(\mu ,F)\varphi _{k}^{\left|\mu \right|}(z)\left|\mu \right|^{n}d\mu . \) 
Notice that
\(\left\Vert \varphi _{k}\right\Vert _{\infty }=c_{n}\frac{(k+n-1)!}{k!}.\) 
So, if we can show that
\(\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}\left|R_{k}\left( \mu ,F\right) \right|\frac{\left( k+n-1\right) !}{k!}\left|\mu \right|^{n}d\mu <\infty \) 
then \(F\)  is bounded. Similarly, if we can show that for any \(N\in \mathbb {N}^{+}\) , there exists a constant \(C_{N}>0\)  such that for each \(j=1,2,\cdots ,m, \)
\(\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}\left|R_{k}\left( \mu ,\left(iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}\left|z\right|^{2}\right) ^{N}F\right) \right|\frac{\left( k+n-1\right) !}{k!}\left|\mu \right|^{n}d\mu <C_{N}\text{,} \) 
then
\(\left|(z,u)\right|^{4N}\left|F\right|^{2}=\left(\sup _{j}\left|u_{j}\right|^{2}+\sup _{j}\frac{\left|\mu _{j}\right|^{2}}{\left|\mu \right|^{2}}\frac{1}{16}\left|z\right|^{4}\right) ^{N}\left|F\right|^{2}=\sup _{j}\left|\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}\left|z\right|^{2}\right) ^{N}F\right|^{2}\le C_{N}^{2},\) 
namely,
\(\left|F(z,u)\right|\le C_{N}\left( 1+\left|(z,u)\right|\right) ^{-2N}.\) 
This yields that for any \(\omega =(x,u)\in \mathbb {G}\)  and \(\mu \in \mathbb {R}^{m}\backslash \lbrace 0\rbrace \) ,
\(\left|S^{\delta }(x,u)\right|=\left|F\left( A_{\frac{\mu }{\left|\mu \right|}}x,u\right) \right|\le C_{N}\left(1+\left|\left( A_{\frac{\mu }{\left|\mu \right|}}x,u\right)\right|\right) ^{-2N}.\) 
Thus, to obtain Theorem REF , it suffices to prove (REF ).
Fixing \(j=1,\cdots ,m\) . We first let \(N=1\)  and calculate \(R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}\left|z\right|^{2}\right) F\right) \) . From (REF ), we know
that
\(R_{k}(\mu ,iu_{j}F)=\frac{2^{(1-n)}k!}{(k+n-1)!}\int _{0}^{\infty }\left(iu_{j}F\right) ^{\mu }(r)\varphi _{k}^{\left|\mu \right|}(r)r^{2n-1}dr.\) 
Since that
\(\left( iu_{j}F\right) ^{\mu }(r)=\frac{\partial }{\partial \mu _{j}}F^{\mu }(r),\) 
then
\(R_{k}(\mu ,iu_{j}F)=\frac{\partial }{\partial \mu _{j}}R_{k}(\mu ,F)-\frac{2^{(1-n)}k!}{(k+n-1)!}\int _{0}^{\infty }F^{\mu }(r)\frac{\partial }{\partial \mu _{j}}\varphi _{k}^{\left|\mu \right|}(r)r^{2n-1}dr.\) 
Noticing
\(\frac{\partial }{\partial \mu _{j}}\varphi _{k}^{\left|\mu \right|}(r)=\frac{1}{2}r^{2}\frac{\mu _{j}}{\left|\mu \right|}\left(L_{k}^{n-1}\right) ^{\prime }(\frac{1}{2}\left|\mu \right|r^{2})e^{-\frac{1}{4}\left|\mu \right|r^{2}}-\frac{1}{4}r^{2}\frac{\mu _{j}}{\left|\mu \right|}L_{k}^{n-1}(\frac{1}{2}\left|\mu \right|r^{2})e^{-\frac{1}{4}\left|\mu \right|r^{2}},\) 
and using the recursion formula (see [1]})
\(r\frac{d}{dr}L_{k}^{n-1}(r)=kL_{k}^{n-1}(r)-(k+n-1)L_{k-1}^{n-1}(r),\) 
we have that
\(\frac{\partial }{\partial \mu _{j}}\varphi _{k}^{\left|\mu \right|}(r)=\frac{\mu _{j}}{\left|\mu \right|}\left( \left|\mu \right|^{-1}k\varphi _{k}^{\left|\mu \right|}(r)-\left|\mu \right|^{-1}(k+n-1)\varphi _{k-1}^{\left|\mu \right|}(r)-\frac{1}{4}r^{2}\varphi _{k}^{\left|\mu \right|}(r)\right) .\) 
Then,
\(R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) F\right) =\frac{\partial }{\partial \mu _{j}}R_{k}(\mu ,F)-\frac{\mu _{j}}{\left|\mu \right|}\left( \frac{k}{\left|\mu \right|}R_{k}(\mu ,F)-\frac{k}{\left|\mu \right|}R_{k-1}(\mu ,F)\right) .\) 
(REF ) and (REF ) yield that
\(R_{k}(\mu ,F)=(1-(2k+n)\left|\mu \right|)_{+}^{\delta }\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|.\) 
We set  \(\sigma =(2k+n)\left|\mu \right|\)  and \(\psi (\sigma )=(1-\sigma )_{+}^{\delta }\) . Then, \(R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) F\right)\)  can be rewritten as
\(&&R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) F\right) \\&=&\frac{\partial }{\partial \mu _{j}}\left( \left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\psi (\sigma )\right) -\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\frac{\mu _{j}}{\left|\mu \right|}\left( \frac{k}{\left|\mu \right|}\psi (\sigma )-\frac{k}{\left|\mu \right|}\psi (\sigma -2\left|\mu \right|)\right) \\&=&\frac{\partial }{\partial \mu _{j}}\left( \left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\psi (\sigma )\right) -\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|k\frac{\mu _{j}}{\left|\mu \right|^{2}}\frac{\partial }{\partial k}\psi \left((2k+n)\left|\mu \right|\right) \\&&+\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|k\frac{\mu _{j}}{\left|\mu \right|^{2}}\left( \frac{\partial }{\partial k}\psi \left( (2k+n)\left|\mu \right|\right) -\psi ((2k+n)\left|\mu \right|)+\psi (\left( 2k-2+n\right) \left|\mu \right|)\right).\) 
Using Taylor expansion, we get that
\(&&\frac{\partial }{\partial k}\psi ((2k+n)\left|\mu \right|)-\psi ((2k+n)\left|\mu \right|)+\psi ((2k-2+n)\left|\mu \right|) \\&=&4\left|\mu \right|^{2}\int _{k-1}^{k}(s+1-k)\psi ^{\prime \prime }((2s+n)\left|\mu \right|)ds.\) 
It follows that
\(&&\int _{\mathbb {R}^{m}}k\frac{\left|\mu _{j}\right|}{\left|\mu \right|^{2}}\left|\frac{\partial }{\partial k}\psi ((2k+n)\left|\mu \right|)-k\psi ((2k+n)\left|\mu \right|)+k\psi ((2k-2+n)\left|\mu \right|)\right|\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\left|\mu \right|^{n}d\mu \\&=&4k\int _{k-1}^{k}(s+1-k)\left( \int _{\mathbb {R}^{m}}\left|\mu _{j}\right|(1-(2s+n)\left|\mu \right|)^{\delta -2}\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\left|\mu \right|^{n}d\mu \right) ds \\&\le &C\int _{k-1}^{k}(2s+n)^{-n-m-1}ds \\&\le &C(2k+n)^{-n-m}\) 
if \((1-\left|\mu \right|)_{+}^{\delta -2}\)  is integrable on \(\mathbb {R}^{m}\) . At the same time, we see that
\(\frac{\partial }{\partial \mu _{j}}\left( \left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\psi (\sigma )\right) -\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|k\frac{\mu _{j}}{\left|\mu \right|^{2}}\frac{\partial }{\partial k}\psi \left((2k+n)\left|\mu \right|\right) =\psi (\sigma )\frac{\partial }{\partial \mu _{j}}\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|.\) 
So,
\(&&\int _{\mathbb {R}^{m}}\left|\frac{\partial }{\partial \mu _{j}}\left(\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\psi (\sigma )\right) -\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|k\frac{\mu _{j}}{\left|\mu \right|^{2}}\frac{\partial }{\partial k}\psi \left( (2k+n)\left|\mu \right|\right)\right|\left|\mu \right|^{n}d\mu \\&=&\int _{\mathbb {R}^{m}}\psi (\sigma )\left|\frac{\partial }{\partial \mu _{j}}\left|\det A_{\frac{\mu }{\left|\mu \right|}}\right|\right|\left|\mu \right|^{n}d\mu \\&\le &\int _{\mathbb {R}}\psi ((2k+n)\lambda )\left( \int _{\mathbb {S}^{m-1}}\left|\frac{\partial }{\partial \eta _{j}}\left|\det A_{\eta }\right|\right|d\eta \right) \lambda ^{n+m-1}d\lambda \\&\le &(2k+n)^{-n-m}.\) 
An iteration of the process shows that \(R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) ^{l}F\right) \) 
also satisfies
\(\int _{\mathbb {R}^{m}}\left|R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) ^{l}F\right)\right|\left|\mu \right|^{n}d\mu \le C_{l}(2k+n)^{-n-m}\) 
provided that \((1-\left|\mu \right|)_{+}^{\delta -2}\)  is
integrable on \(\mathbb {R}^{m}\) . Thus, when \(l=N\)  and \(\delta >2N-1\)  we have
\(\int _{\mathbb {R}^{m}}\left|R_{k}\left( \mu ,\left( iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}r^{2}\right) ^{N}F\right)\right|\left|\mu \right|^{n}d\mu \le C_{N}(2k+n)^{-n-m},\) 
which implies that
\(\sum _{k=0}^{\infty }\int _{\mathbb {R}^{m}}\left|R_{k}\left( \mu ,\left(iu_{j}-\frac{\mu _{j}}{\left|\mu \right|}\frac{1}{4}\left|z\right|^{2}\right) ^{N}F\right) \right|\frac{\left( k+n-1\right) !}{k!}\left|\mu \right|^{n}d\mu <\infty .\) 
The proof of Theorem REF  is completed.
